Marisa M Souza
AP Biology
Mrs.Therrien
October, 6, 2015 Diffusion and Osmosis
Exercise 1A: Diffusion In our diffusion experiment glucose is leaving the bag and Iodine-Potassium-Iodide or IKI is entering the bag. The change in color within the bag indicates the presence of glucose. The IKI moved from the beaker into the dialysis bag, this resulted in a change of color to black within the bag. This made the concentration within the bag equal to the concentration outside the bag. The glucose solution moved out of the bag making glucose present in the beaker. The glucose moved to make the solute concentration inside and out equal. If the initial and final % concentration of glucose and IKI for in the bag and in the beaker were given they …show more content…
In this experiment we tested the theory that water will always move from an area of higher water potential to an area of lower water potential. Water potential is affected by both pressure potential and solute potential. Our resulting water potential will be negative because pure water has a water potential of 0 and the addition of any solute would only further decrease it. C was calculated by determining the equation for the line of best fit which was roughly y=-38x+7, where y=0 (the x-axis on our graph, Refer to The Effect of Sucrose Concentrations on Russet Potatoes through 24 hour Immersion graph). For our class data the solute potential of the russet potatoes was -4.618 bars. This was calculated using the equation () = –iCRT, where i is the ionization constant 1, C is the molar concentration, which was determined to be .189, R was our pressure constant which is .08311 liter bars/moles kelvin, and T was the temperature which was 294 Kelvin in the room. Because the pressure potential of the solution is zero, our solute potential is equal to the water potential. The water potential of the solution at equilibrium will be equal to the water potential of the potato cells. This value is useful because it allows us to predict the direction of the flow of …show more content…
Our team tested blue potatoes and as expected the results were similar yet different. Our C was calculated the same way as in part 1C, the equation for our line of best fit was approximately y=-40x+14. We then plugged it into the equation () = –iCRT, where i is the ionization constant 1, C is the molar concentration, which was determined to be .344, R was our pressure constant which is .08311 liter bars/moles kelvin, and T was the temperature which was 294 Kelvin in the room. This resulted in a solute potential of -8.408 bars. Because the pressure potential of the solution is zero, our solute potential is equal to the water potential. The water potential of the solution at equilibrium will be equal to the water potential of the potato cells. This value is useful because it allows us to predict the direction of the flow of water. The lower water potential of the blue potato indicates that if the russet and the blue potatoes were put in the same solution water would be more likely to flow into the blue potato. Other groups tested different types of potatoes in this part the group with sweet potatoes resulted in a water potential of -11.73 bars, the only water potential rendered that was higher than ours, the white harvest potato resulted in a water potential of -4.153 bars very similar to that of the russet.The distribution of points on